HDU2066

    现在终于知道什么是水题，模板还没理清，但一套便过。轻轻的我一飘过，这道题便AC了。

 #include <iostream>


#include <cstdio>

#include <cstring>

#include <algorithm>

#include <map>

#include <vector>

#include <queue>

#include <stack>

#define maxn 100005

#define INF 999999999

using namespace std;

struct edge

{

   int to;

   int cost;

   int from;

};







int V,E;                //V:顶点数，E：边数。

edge es[maxn];

//vector<edge>G[maxn];

int d[maxn];

void shortest_path(int s)         //传统Bellman-ford打法。

{

  for(int i = 1;i <=V; i++)

    d[i] = INF;              //初始化

  

  d[s] = 0;

  while(true)

  {

    bool update = false;

    for(int i = 1; i <= 2*E; i ++)

    {

      edge e = es[i];

      //printf("%d %d %d\n",e.from, e.to, e.cost);

      //printf("%d %d \n",d[e.to],d[e.from]);

      if(d[e.from] != INF && d[e.to] > d[e.from]+e.cost)

      {

        d[e.to] = d[e.from]+e.cost;

        //printf("%d\n",e.cost);

        update = true;

      }

    }

    if(!update)break;

  }

}

int main()

{

    int i, a,j;

    int T, S, D;

    int min1;

    int p[1000],q[100000] ;

    edge b;

    while(~scanf("%d%d%d",&T,&S,&D))

    {

       //if(V==0&&E==0)break;

       for(i= 1,V = 0;i <= 2*T; i++)

       {

         scanf("%d%d%d",&es[i].from, &es[i].to, &es[i].cost);

         i++;

         es[i].from = es[i-1].to;

         es[i].to = es[i-1].from;

         es[i].cost = es[i-1].cost;

         if(V<es[i].from)V = es[i].from;

         if(V<es[i].to)V = es[i].to;

         //G[a].push_back(b);

         //printf("%d %d",b.to,b.cost);  

       }

       E = T ;

       for(i = 1; i <= S; i ++)

         scanf("%d",&p[i]);

         

       for(i = 1; i<= D; i++)

        scanf("%d", &q[i]);

       

       for(i = 1,min1 = INF; i <= S; i ++)            //启动多重单源最短路模式。。此题比较水。。居然没超时。。

       {

         shortest_path(p[i]);

         for(j = 1; j <= D; j ++)

         {

            if( d[q[j]] < min1)min1 = d[q[j]];

         }

       }

       

       printf("%d\n", min1);

      }

    return 0;

}